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Tickets cost $9 for students and $15 for adults, 420 tickets were sold, and the takings came to $4,860. Which pair of rules matches this situation?
As + a = 4860 and 9s + 15a = 420
B9s + 15a = 420 and s + a = 4860
Cs + a = 420 and 9s + 15a = 4860
Ds + a = 420 and 15s + 9a = 4860
Answer & Solution
Correct answer: C. s + a = 420 and 9s + 15a = 4860
1. One rule must count tickets, and 420 is the number of tickets, so s + a = 420.
2. The other rule must add up money, so each count is multiplied by its own price.
3. Student tickets bring in 9s dollars and adult tickets bring in 15a dollars, giving 9s + 15a = 4860.
4. Options A and B swap the two totals, so tickets are counted in dollars and money in tickets.
5. Option D attaches each price to the wrong group, which would make adults the cheaper ticket.
_Source: OpenStax College Algebra (CC BY 4.0), Ch 7 "Systems of Equations and Inequalities", section 7.1 Systems of Linear Equations: Two Variables_
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